Isomorphism from <R,+> to <R+,\times>: Proving 1-1 and Onto Function

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In summary, the conversation is about determining whether there is an isomorphism from <R,+> to <R+,\times> where \phi(r)=0.5^{r} when r \in R. The speaker asks about using logarithmic functions to show that the function is 1-1, and another speaker suggests rewriting the equation to make it cleaner. The speaker also asks about whether proving 1-1 and onto is enough to show an isomorphism, and the other speaker reminds them that additional properties must be proven as well.
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LMKIYHAQ
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Homework Statement



Is there an isomorphism from <R,+> to <R+,[tex]\times[/tex]> where [tex]\phi[/tex](r)=0.5[tex]^{r}[/tex] when r [tex]\in[/tex] R?

2. Homework Equations
For an isomorphism I know it is necessary to show there is a 1-1 and onto function. I am unsure if I can use the steps I am trying to use to show it is 1-1.

The Attempt at a Solution



For phi(r)=phi(s) I want to show r=s. Am I able to take the ln (or log?) of both sides to get ln(.05[tex]^{r}[/tex])=ln(0.5[tex]^{s}[/tex])? I am not sure which to use (ln or log) and where these logarithmic functions would be defined since for r=s, r and s are supposed to be real numbers.

Thanks for the help.
 
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  • #2
Maybe try rewriting 0.5^r=0.5^s as 2^r=2^s. This may make things cleaner.
 
  • #3
I don't know how to make that change?
 
  • #4
Even if that way does clean it up, is my way of taking ln of both sides wrong?
 
  • #5
Sure, just use logs. (0.5)^r=(0.5)^s iff r*log(0.5)=s*log(0.5). That shows it 1-1. Is it onto? But 1-1 and onto doesn't make it an isomorphism. You have to prove things like phi(r+s)=phi(r)*phi(s), right?
 
  • #6
Thanks for the help.
 

1. What is an isomorphism?

An isomorphism is a mathematical concept that refers to a bijective function between two mathematical structures that preserves the structure of the two structures. In the context of this question, it refers to a bijective function between the additive group of real numbers and the multiplicative group of positive real numbers .

2. How do you prove that the function is 1-1?

To prove that a function is 1-1 (injective), we need to show that for any two distinct elements in the domain, their images in the codomain are also distinct. In other words, each element in the domain is mapped to a unique element in the codomain. In the case of this isomorphism, we need to show that for any two distinct real numbers, their sum will always be different from their product.

3. How do you prove that the function is onto?

To prove that a function is onto (surjective), we need to show that every element in the codomain has at least one preimage in the domain. In other words, every element in the codomain is mapped to by at least one element in the domain. In the case of this isomorphism, we need to show that for every positive real number, there exists a real number whose sum with another real number will equal the positive real number.

4. What are the steps to proving an isomorphism between two groups?

The steps to proving an isomorphism between two groups are as follows:

  • Show that the function is bijective (1-1 and onto).
  • Show that the function preserves the group operation. In other words, if a and b are elements in the domain, then f(a+b) = f(a) * f(b) in the codomain.

5. Why is it important to prove that the function is an isomorphism?

Proving that a function is an isomorphism is important because it shows that two mathematical structures are essentially the same, even though they may appear different. It also allows us to use properties and theorems from one structure in the other structure, which can be useful in solving problems and making connections between different areas of mathematics.

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